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I need to evaluate $$\int_0^1\frac{\sin^{-1}x}x dx.$$

I have tried integration by parts (taking $1/x$ as second function) but I am getting complex results. Any idea on how to approach will be appreciated.

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  • $\begingroup$ -1 stands for inverse $\endgroup$ – user562886 May 18 '18 at 3:51
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This is tricky. Do integration by parts (assuming you justify convergence at $0$). Then you're left with $$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx.$$ Substitute $x=\sin u$ and note that you can do the definite integral $$\int_0^{\pi/2} \ln(\sin u)\,du = \frac12\int_0^\pi \ln(\sin u)\,du$$ by the trick of substituting $u=2v$ and using the double angle formula and symmetry.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x & = -\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x = -\,{1 \over 4}\int_{0}^{1}x^{-1/2}\ln\pars{x}\pars{1 - x}^{-1/2}\,\dd x \\[5mm] & = \left. -\,{1 \over 4}\,\partiald{}{\nu}\int_{0}^{1}x^{\nu -1/2} \,\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ \nu\ =\ 0} \\[5mm] & = -\,{1 \over 4}\,\partiald{}{\nu}\bracks{\Gamma\pars{\nu + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{\nu + 1}}_{\ \nu\ =\ 0} \\[5mm] & = -\,{1 \over 4}\,\root{\pi}\,\partiald{}{\nu}\bracks{\Gamma\pars{\nu + 1/2} \over \Gamma\pars{\nu + 1}}_{\ \nu\ =\ 0} \\[5mm] & = \bbx{{1 \over 2}\,\ln\pars{2}\,\pi} \approx 1.0888 \end{align}

Note that $\ds{\Gamma\pars{\nu + a} \sim \Gamma\pars{a} + \Gamma\pars{a}\Psi\pars{a}\nu}$ as $\ds{\nu \to 0}$ such that

\begin{align} &{\Gamma\pars{\nu + 1/2} \over \Gamma\pars{\nu + 1}} \sim {\Gamma\pars{1/2} \over \Gamma\pars{1}}\, {1 + \Psi\pars{1/2}\nu \over 1 + \Psi\pars{1}\nu} \sim \root{\pi}\bracks{1 + \Psi\pars{1 \over 2}\nu} \bracks{1 - \Psi\pars{1}\nu} \\[5mm] \sim &\ \root{\pi}\braces{\rule{0pt}{5mm}1 + \bracks{\Psi\pars{1 \over 2} - \Psi\pars{1}}\nu} \quad \mbox{as}\ \nu \to 0 \\[5mm] &\ \mbox{and}\ \Psi\pars{1 \over 2} - \Psi\pars{1} = \int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t = 2\int_{0}^{1}{t - 1 \over 1 - t^{2}}\,\dd t = -2\int_{0}^{1}{\dd t \over 1 + t}\,\dd t = -2\ln\pars{2} \\[5mm] &\ \mbox{such that}\ \pars{-\,{1 \over 4}\,\root{\pi}}\root{\pi}\bracks{-2\ln\pars{2}} = \bbx{{1 \over 2}\,\ln\pars{2}\,\pi} \end{align}

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Just another way.

Consider $$\sin ^{-1}(x)= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}$$ $$\int_0^a\frac{\sin^{-1}(x)}x\, dx=\sum_{n=0}^\infty \int_0^a \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n}}{2n+1}\,dx=a \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};a^2\right )$$ where appears the hypergeometric function which beautifully simplies for $a=1$ (only).

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